Volumes
Volume of a solid sphere hole A round hole of ra...

Question

Volumes

Volume of a solid sphere hole A round hole of radius √3 ft is bored through the center of a solid sphere of radius 2 ft. Find the volume of material removed from the sphere.

Accepted Answer

Hello there, today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A cylindrical hole of radius 3 multiplied by the square root of 3 units is bored through the center of a solid sphere of radius 6 units. Determine the volume of material removed from the sphere. So it appears for this particular prompt, we are given a solid sphere, so let's draw a sphere so we can help visualize this prompt. So we have a sphere, specifically a solid sphere. And in the sphere, we have a hole that is bored through the center, so that means you'll literally have a cylindrical shape that goes straight through to the other side of our sphere. And this cylindrical hole, so meaning just imagine this long, you know, cylinder that's removed from the solid sphere. The cylinder has the cylinder shape has a radius of 3 multiplied by the square root of 3 units. And the radius of our solid sphere itself is 6 units, and based on this information, our final answer that we're ultimately trying to solve for is we're asked to determine what the volume of the material removed from the sphere will be. So what is the volume of the cylindrical shape that is removed from our solid sphere? So with that in mind, our first step that we need to take is we need to recall and use our information about cylindrical shells, noting that the radius of our cylindrical shell will be lowercase r, and the thickness will be dr, and using our information about cylindrical shells, we will know that they are centered on the axis of the board hole. So for the condition, 0 is less than or equal to lowercase r, which is less than or equal to a. That will mean that each shell will have a circumference c that is equal to 2 multiplied by pi multiplied by r, and they will have a height that is equal to the cord of each sphere at a distance lower case r from the axis, namely the square root of. Or specifically, namely, 2 multiplied by the square root of R to the 2 minus lowercase r to the 2, where this value of R is the radius of the sphere. Thus, the expression or the formula for the volume that is removed from the sphere will be V the volume is equal to the integral evaluated from 0 to a of 2 multiplied by pi multiplied by r multiplied by parentheses of 2 multiplied by the square root of R 2 minus lowercase r 2. The r, which will be equal to 4 pi multiplied by the integral evaluate from 0 to a of lowercase r multiplied by the square root of R to the 2 minus r to the power of 2. DR fantastic. And we now, at this stage, we need to let A be equal to 3 multiplied by the square root of 3, and capital R be equal to 6. And when we plug in our known variables, we will find that the volume V will be equal to 4 pi multiplied by the integral, evaluated from 0 to 3 multiplied by the square root of 3. Of lowercase r multiplied by 62 minus r to 2, the r, which is equal to 4 pi multiplied by the integral, evaluate from 0 to 3 multiplied by the square of 3 of lowercase r multiplied by the square root of 36 minus lowercase r to the power of 2. DR Now we need to use our substitution method. So recalling and using our substitution method, we will state that you will be equal to 36 minus lowercase r to 2. Then that will mean, as a result, that we can go ahead and write that DU by DR will be equal to -2R, which will mean that DU will be equal to 2 RDR. So now at this stage, we need to rewrite our limits. So when R is equal to 0, that will mean that you will be equal to 36 minus 0 to 2. Which will be simply equal to 36, and when r is equal to 3 multiplied by the square root of 3, that will mean that u is equal to 36 minus parentheses 3 multiplied by the square root of 3 to the 2, which will be equal to 36 minus 27, which will be simply equal to 9. So, therefore, the volume V will be equal to 2 pi multiplied by this integral of 0 to the, so it's gonna be evaluated from 0 to 3 multiplied by the square root of 3 of the square root of 36 minus lowercase r to the 2 multiplied by parentheses of 2 RDR. Which will mean we can write that our volume V will be equal to -2 pi multiplied by the integral, evaluated from 36 to 9 of u to the power of 1/2 du. Which we can write that V, the volume will be equal to 2 pi multiplied by the integral, evaluate from 9 to 36 of U to the power of 1/2 DU. Which will mean that V, the volume will be equal to 2 pi multiplied by u to the power of 3 divided by 2 divided by 3 divided by 2. Evaluate from 9 to 36. Which will be equal to or pi divided by 3 multiplied by u multiplied by the square root of u evaluated from 9 to 36, which will be equal to or pi divided by 3 multiplied by square brackets of 36 multiplied by the square root of 36 minus. 9 multiplied by the square root of 9. And when we plug this expression into our calculator and we write our final answer in its most simplified form, we will find that it's simply equal to 252 pi, and its units will be cubic units. And that's it. That is our final answer. Hooray, we did it. So going back to the top to recap what we've learned, we can safely say that our final answer without a doubt. For this particular problem for the volume will be 252 pi cubic units, and that's it, that is our final answer. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

VolumesVolume of a solid sphere hole A round hole of radius √3 ft is bored through the center of a solid sphere of radius 2 ft. Find the volume of material removed from the sphere.

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Volumes
Volume of a solid sphere hole A round hole of radius √3 ft is bored through the center of a solid sphere of radius 2 ft. Find the volume of material removed from the sphere.